TWO-PARAMETER TAXICAB TRIG FUNCTIONS

KELLY DELP AND MICHAEL FILIPSKI

Abstract. In this paper, we review some of the fundamental properties ofthe `1, or taxicab, metric on R

2. We define and give explicit formulas for two-parameter sine and cosine functions for this metric space. We also determinethe maximum of these functions, which is greater than 1.

1. Introduction

The `1 metric on R

2, the so-called taxicab metric, is often one of the first non-Euclidean metrics a mathematics student encounters. For any points p = (p1, p2)and q = (q1, q2) in R

2, the metric is given by the formula

dT (p, q)=|p1 � q1 |+ |p2 � q2 | .The `1 metric is just one metric in a class of metrics defined on R

2 known asMinkowski metrics; see [3] for an introduction to these spaces. Let ⌦ be a closed,bounded convex set in R

2 which contains and is symmetric about the origin. Theset ⌦ defines a norm on R

2, where ⌦ is the unit disk. Given a norm k · k, one candefine a metric on R

2 by d(p, q) = kp� qk. Examples of Minkowski metrics includethe `p metrics, the `1 or max metric, and metrics with unit disk a regular 2n-gon.

Length minimizing paths in the taxi-cab plane are not necessarily unique, so weuse the vector space properties of R2 and define lines to be the sets of points ofthe form L = {tv+b | t 2 R} for some fixed v and b. We can similarly define line

segments, triangles, rays, and angles (pairs of rays sharing an initial point). Wedefine the length of a line segment AB to be the distance between the endpoints,dT (A,B).

Given a metric d on a set X, a circle C of radius r is the set of all points p 2 Xequidistant from a given point called the center. A circle in the taxicab metricis a square with diagonals parallel to the x and y-axis. In Euclidean space thereis an intrinsic notion of angle measure, radian measure, which is determined bythe length of an a particular circle arc. We can similarly define an intrinsic anglemeasure in the taxicab plane, called t-radians.

Definition 1. Let C be a circle with radius r, and center P . Given an angle with

vertex P , let s be the length of the subtended arc. The t-radian measure, ✓, of ataxicab angle, is given by

✓ =s

r.

It is this notion of angle measure which was used in these previous works [1],[5], and [8], on taxicab trigonometry. Another well-studied angle measure in aMinkowski metric uses the area of the sector of the circle, rather than arc length,to define the angle measure. (Due to a theorem of Haar, any area measure µis proportional to Lebesgue measure; see [4] for a discussion of areas in normed

1

2 KELLY DELP AND MICHAEL FILIPSKI

spaces.) By Theorem 1 in [9], these two notions are equivalent (up to scale) becausethe taxicab circle is an example of an equiframed curve. See [9] for the definitionof equiframed curve.

Note that an `1 circle has 8 t-radians, which means in this metric, 4 is theanalogue of ⇡. Some of the properties from Euclidean geometry have analogousstatements which are true in the taxicab plane. We will use the following proposi-tions, both of which can be found in [8].

Proposition 1 (Theorem 4.2 [8]). The angle sum of a taxicab triangle is 4 t-

radians.

We define a taxicab right angle to be an angle with measure 2 t-radians, which,as in Euclidean geometry, is an angle which has measure equal to its supplement.

Proposition 2 (Lemma 2 [8]). A Euclidean right angle has taxicab angle measure

of 2 t-radians, and conversely.

Figure 1 gives a sketch of a proof of Proposition 2.

x

x

2-x

Figure 1. Euclidean right angles have taxicab angle measure of 2.

Proposition 2 implies that the vectors x and y form a right angle in the taxi-cab plane if and only if they are orthogonal in the Euclidean sense. The study ofdi↵erent notions of orthogonality in Minkowski spaces is an active area of research.Two important orthogonality types in Minkowski spaces are Birkho↵ orthogonality,(x ? y if and only if kx� ↵yk � kxk for all ↵) and James/isosceles orthogonality(x ? y if and only if kx+yk = kx-yk.) In the taxicab plane, Birkho↵ orthogonalityis not symmetric, and James orthogonality is not invariant under scaler multiplica-tion of the vectors, which implies neither are equivalent to definition of right anglethat we use above; see the recent survey article [2] for an explanation of these factsand extensive discussion of orthogonality in normed linear spaces.

Not all angles in the taxicab geometry behave as nicely as right angles. In Figure2, the Euclidean angles ↵ and � of the two triangles depicted are not equal, butthe taxicab angle measure of both is 1

2 .A taxicab right triangle is in standard position if the base of the triangle is parallel

to the x-axis (see ↵-triangle in Figure 2). For triangles in standard position, we

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 3

can define the taxicab sine and cosine functions as we do in Euclidean geometrywith the cos ✓ and sin ✓ equal to the x and y-coordinates on the unit circle. Indeed,the piecewise linear formulas for these functions are given in [8] and [1], and withslightly di↵erent formulas in [5]. However, if we define the sine and cosine as ratioof sides of right triangles, considering only triangles in standard position will notgive all possible values. To illustrate this, we refer again to Figure 2.

↵

�

Figure 2. An `1 circle with two right-angled triangles.

Both triangles are right triangles with hypotenuse (the side opposite the 2 t-radian angle) of length 1. Also, since ↵ and � both have angle measure 1

2 , the othernon-right angle is 4� 2� 1

2 = 32 . In the ↵-triangle, we compute the cosine of ↵ by

taking the ratio of the lengths of the adjacent side to the hypotenuse, which is 34 .

However, looking at the �-triangle, we see the vertex of the right angle falls outsideof the unit circle, which implies the length of the side adjacent to �, and thereforethe cosine of �, is greater than one.

A natural question arises: what is the maximum value of the cosine of an anglein the taxi-cab plane? In this paper, we define and give explicit formulas for two-parameter sine and cosine functions, describing the possible side ratios of right-triangles in the taxi-cab plane. Using these formulas we show the maximum valueto be 1

2 +1p2, which is greater than 1. Thus we obtain a quantitative measure of a

di↵erence between the Euclidean and taxi-cab plane.We would like to thank the referee for pointing out many references on the

geometry of Minkowski metric spaces, including [7]. In Chapter 8 of this text,Thompson defines two-parameter sine and cosine functions for general Minkowskispaces. For Thompson’s function, the Minkowski cosine of two vectors is zero ifand only if the vectors x1 and x2 are Birkho↵ orthogonal. This property does nothold for our definition of cosine, so our functions are not a special case of thosedefined by Thomspon, even up to scale. Using the sine function, Thompson definesan ↵ which measures how far the Minkowski space is from Euclidean space, leavingus with a question: Is this ↵ related to the value we obtain for maximum of ourtaxicab sine function?

4 KELLY DELP AND MICHAEL FILIPSKI

2. A two-parameter sine and cosine function

Definition 2. Given two metric spaces (X, d1) and (Y, d2), a bijection f : X ! Yis an isometry if for any two points p, q 2 X:

d1(p, q)=d2(f(p), f(q))

Given a metric space X, the set of all isometries � : X ! X forms a group, andthe set of isometries that fix a point forms a subgroup of this group. An importantsubgroup is the set of isometries which fix the origin, which, by the Mazur-UlamTheorem (see [7], Chapter 3), are linear. Using this fact and the fact that isometriesmap circles to circles with the same radius, one can see that the group of isometriesthat fix the origin of (R2, dT ) is the group of symmetries of a square, also called thedihedral group D4. This includes the set of rotations (by 90�, 180�, and 270�) andreflections across the x-axis, y-axis and the lines passing through the origin withslope ±1. The full group of isometries is the semi-direct product D4 o R

2, whichis proven in [6]. This group is generated by translations and isometries that fix theorigin.

Two triangles T1, T2 in the taxicab plane are congruent if there is a taxicabisometry � such that �(T1) = T2. Note that due to the rigidity of the isometrygroup, there is no taxicab isometry taking the ↵-triangle in Figure 2 to the �-triangle, so there is no angle-side-angle theorem in taxicab geometry. We willdefine the taxicab sine and cosine functions to have two angle parameters; oneparameter is the usual ✓-angle parameter measured from a fixed axis, and the other�-parameter will denote the “direction” of the triangle in the plane (see Figure 3).

Before making the definition, we describe a notion of orthogonal projection inthe taxicab plane. Let L be a line and P be a point. If P is on L, the orthogonal

projection of P onto L is P . If P is not on L, the orthogonal projection is a uniquepoint R on L for which the line segment PR makes a Euclidean right angle with L;Proposition 2 implies that this point R is also the unique point on L which makesa taxicab right angle. The following definition, which is convenient for later proofs,may seem somewhat unnatural; we refer the reader to Propositions 3 and 4 whichjustify that this definition gives the desired “signed ratio” of side lengths.

Definition 3. Let L be the line through the origin O which makes reference angle

0 � < 2 with the x-axis, and let P = (p1, p2) be a point on the unit circle so that

OP makes angle ✓ with L. Let R = (r1, r2) be the orthogonal projection of P onto

L. We define the taxicab cosine and sine of angle ✓ at reference angle � as:

tcos�(✓) = r1 + r2 tsin�(✓) = (r1 � p1) + (p2 � r2)

Given a right triangle T with hypotenuse equal to one, there is a taxi-cab isom-etry which maps T to a triangle of the form �PRO given in Definition 3, so T iscongruent to �PRO.

Let L? be the perpendicular to L which also passes through the origin. The linesL and L? divide the plane into four quadrants, which we will refer to by numberingcounterclockwise I, II, III, and IV.

Proposition 3. The value of tcos�(✓) is positive for ✓ in quadrants I and IV,

and negative for ✓ in quadrants II and III. Similarly tsin�(✓) is positive for ✓ in

quadrants I and II, and negative for ✓ in quadrants III and IV.

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 5

�✓

L?

L

(r1, r2)

(p1, p2)

Q

Figure 3. Defining sine and cosine.

Proof. Let P = (p1, p2) and R = (r1, r2) be as given in Definition 3. When ✓ is inquadrants I and IV, as defined by L and L?, the coordinate r1 is positive and r2 isnon-negative (when � = 0, the line L is the x-axis and r2 = 0). Therefore tcos�✓,which is the sum of these coordinates, is positive. Similarly when ✓ is in quadrantsII and III, r1 is negative and r2 is non-positive, hence tcos�✓ is negative.

Recall that tsin�✓ = (r1 � p1) + (p2 � r2). For fixed �, the coordinates of P andR are continuous real-valued functions of ✓, and therefore the functions r1�p1 andp2 � r2 are also continuous functions. When 0 < � < 2, each of these functions iszero if and only if ✓ = 4n for some integer n. This follows from the fact that theslope of L is positive, which implies the line through P and R has negative slope,so p1 = r1 or p2 = r2 if and only if P = R. Therefore the sign of each of thesefunctions, r1 � p1 and p2 � r2, is constant for ✓ in quadrants I and II. Picking aspecific angle such as ✓ = 2 allows us to verify that both are positive, and thereforetsin�✓ is positive. Choosing an angle in the range 4 < ✓ < 8 shows that bothof these functions are negative, and therefore tsin�✓ is also negative when ✓ is inquadrants III and IV.

When � = 0, we have r2 = 0, and r1 = p1, so tsin�✓ = p2, and the result follows.⇤

Proposition 4. In the right triangle made by P , R and the origin O, |tcos�(✓)|gives the length of the adjacent side to ✓, and |tsin�(✓)| gives the length of the

opposite side.

Proof. Fix an angle 0 � < 2. The length of the adjacent side is the distance fromR to the origin, which is |r1| + |r2|. When ✓ is in quadrants I and IV (defined byL and L?), both r1 and r2 are non-negative, so

|r1|+ |r2| = r1 + r2 = |tcos�(✓)|.When ✓ lies in quadrants II and III, both r1 and r2 are non-positive, so

6 KELLY DELP AND MICHAEL FILIPSKI

|r1|+ |r2| = �r1 � r2 = �(r1 + r2) = |tcos�(✓)|.The length of the opposite side of ✓ in triangle OPR is given by the distance

between P and R, which is |p1 � r1|+ |p2 � r2|. Arguing as in Proposition 3, when✓ is in quadrants I and II we have

|p1 � r1|+ |p2 � r2| = (r1 � p1) + (p2 � r2) = |tsin�(✓)|,and when ✓ is in quadrants III and IV,

|p1� r1|+ |p2� r2| = �(r1� p1)� (p2� r2) = � [(r1 � p1) + (p2 � r2)] = |tsin�(✓)|.

⇤

Proposition 5. The following identities hold.

tsin�(✓ � 4) = �tsin�(✓) and tcos�(✓ � 4) = �tcos�(✓).

Proof. Let P and R be the points given in Definition 3 corresponding to ✓, andP 0 and R0 the points corresponding to ✓ � 4. By Proposition 2, taxi-cab angles ofmeasure 2 are Euclidean right angles, which means P and P 0 are antipodal pointson the unit circle and P 0 = �P . The map (x, y) ! (�x,�y) is an isometry of thetaxi-cab plane, which maps P to P 0. Angles are defined by the metric, thereforeisometries preserve angle measure. It follows from the definition of R that R0 = �R.Therefore,

tcos�(✓ � 4) = �r1 � r2 = �(r1 + r2) = �tcos�(✓)

and

tsin�(✓ � 4) = (�r1 + p1) + (�p2 + r2) = � [(r1 � p1) + (p2 � r2)] = �tsin�(✓).

⇤

3. Explicit formulas for Sine and Cosine functions

Theorem 1. Let � be a taxicab reference angle such that 0 � < 2 and let ✓ be a

taxicab angle measured relative �. Let ↵ =1

�2 � 2�+ 2, which is well defined for

all �, since �2 � 2� + 2 > 0. The sine and cosine of ✓ with reference angle � are

given by:

tsin�✓ =

8>>>>>>>><

>>>>>>>>:

↵ ✓ �� ✓ 2� �

1 + ↵(✓ � 2)(�� 1) 2� � ✓ 4� �

↵(4� ✓) 4� � ✓ 6� �

�1 + ↵(6� ✓)(�� 1) 6� � ✓ 8� �

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 7

tcos�✓ =

8>>>>>>>><

>>>>>>>>:

1 + ↵ ✓(�� 1) �� ✓ 2� �

↵(2� ✓) 2� � ✓ 4� �

�1 + ↵(4� ✓)(�� 1) 4� � ✓ 6� �

↵(✓ � 6) 6� � ✓ 8� �

Lemma 1. Let L be a line through the origin that makes angle � with the x-axissuch that 0 � < 2. The point of intersection between L and the unit taxicab circle

is Q =

✓2� �

2,�

2

◆.

Proof. Let Q = (q1, q2). Since Q lies on the unit circle and 0 � < 2, bothcoordinates are positive and

(1) q1 + q2 = 1.

Since the radius of the unit circle is 1, the definition of angle implies that � is thedistance between Q and (1, 0). This distance is given by:

(2) |q1 � 1|+ |q2 � 0| = 1� q1 + q2 = �.

We solve the system of linear equations consisting of (1) and (2) for q2 by addingthe two equations to get

q2 =�

2;

substituting q2 into (1) gives us q1 = 1� �

2, which is the desired result. ⇤

3.1. Proof of Theorem 1 for �� ✓ 2� �.

Proof. Let 0 � < 2, and �� ✓ 2 � �. We will determine the coordinates ofP and R, given in Definition 3, as functions of � and ✓. Lemma 1 implies that the�-axis (line L in Figure 3) intersects the circle at

Q =

✓2� �

2,�

2

◆.

Since the �-axis passes through the origin, we find that the equation is:

(3) L(x) =�

2� �x.

Next, we determine the coordinates of P , the point of intersection between thecircle and the (✓ + �)-ray. Applying Lemma 1 again with angle (✓ + �) givescoordinates:

P =

✓2� �� ✓

2,�+ ✓

2

◆.

Proposition 2 implies that Euclidean right angles are taxi-cab right angles.Therefore, to find the point R we determine the equation of the line perpendic-ular (in the usual Euclidean sense) to the �-axis, LP , through point P . Since the

8 KELLY DELP AND MICHAEL FILIPSKI

�-axis has slope�

2� �, LP has slope

�� 2

�. Since we know the coordinates of

P = (p1, p2) and the slope, we can determine the equation for LP , which is

LP (x) =

✓�� 2

�

◆(x� p1) + p2

=(�� 2)x

�+

(�� 2)(✓ + �� 2) + �(✓ + �)

2�.(4)

The point R is the intersection between the �-axis and LP . Setting equations 3and 4 equal to each other and solving for the x-coordinate of R yields

r1 =2� �

2+

(2� �)(�✓ � ✓)

2(�2 � 2�+ 2).

Plugging r1 into L (or Lp) gives the y-coordinate of R,

r2 =�

2+

�2✓ � �✓

2(�2 � 2�+ 2).

Thus, the coordinates of R are:

R =

✓2� �

2+

(2� �)(�✓ � ✓)

2(�2 � 2�+ 2),�

2+

�2✓ � �✓

2(�2 � 2�+ 2)

◆.

The result now follows by using the coordinates of R and P to compute tsin�(✓)and tcos�(✓) by the formulas given in Definition 3. ⇤

3.2. Proof for 2� � ✓ 4� �.

Proof. We again find the coordinates of P and R to compute tsin�(✓) and tcos�(✓).When 2 < ✓ + � < 4, the point P is in the second quadrant (as defined by the x-and y-axes). Let ✓1 be the portion of ✓ measured from the y-axis, so ✓1 = �+✓�2.

Let f : R2 ! R

2 be the map defined by (x, y) 7! (y,�x). This map is an order4 isometry of the `1 metric. Note that f(0, 1) = (1, 0) and f(P ) is in the firstquadrant. Since angle measure is defined by the metric, angle measure is preservedby isometries. We can therefore apply Lemma 1 to f(P ) to obtain the coordinates

f(P ) =

✓2� ✓1

2,✓12

◆.

To obtain the coordinates for P we apply the inverse map:

P = f�1

✓2� ✓1

2,✓12

◆=

✓�✓1

2,2� ✓1

2

◆=

✓2� �� ✓

2,4� �� ✓

2

◆.

To finish the proof for this interval, we use the same procedure as in the prooffor the first interval; that is, we find the equation of the line perpendicular to the�-axis through P to determine the coordinates of the point R. The line through Pperpendicular to L(x) is

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 9

LP (x) =

✓�� 2

�

◆(x� p1) + p2

=(�� 2)x

�+

(�� 2)(✓ + �� 2) + �(4� ✓ � �)

2�.(5)

To find r1, we set equations 3 and 5 equal to one another and solve for x, whichgives:

r1 =(�� 2)(✓ � 2)

2(�2 � 2�+ 2).

Plugging r1 into L(x) (equation 3) gives r2:

r2 =��(✓ � 2)

2(�2 � 2�+ 2).

The sine and cosine functions can now be computed from the formulas given inDefinition 3.

⇤

3.3. Proof for 4� � ✓ 8� �.

Proof. We will use the symmetry of the functions to establish the formulas for thethird and fourth intervals. Let ✓ be in the given interval, and ✓⇤ = ✓ � 4. Then�� ✓⇤ 4� �. We have determined formulas for tsin�(✓⇤) and tcos�(✓⇤) in thisinterval, so applying Proposition 5 gives formulas for angle ✓ in the remaining twointervals.

⇤

It should be noted that our formulas are a generalization of those formulas in [8]and [1]; if � = 0, then ✓ is in standard position and we obtain identical formulas.

4. Properties of the Functions

4.1. Periodic Extensions and Graphs. In Definition 3, the generalized sine andcosine functions were defined for all real numbers ✓ and for values of � such that0 � < 2. It is evident from the definition that the ✓-period of these functions is8, so for any integer k,

tcos�(✓ + 8k) = tcos�(✓) and tsin�(✓ + 8k) = tsin�(✓).

There is a natural �-extension of these functions; since rotation by right anglesgives isometries of the `1 metric, we extend the �-domain of the generalized sineand cosine functions to be �-periodic with period 2. Therefore, for any integer s,

tcos�+2s(✓) = tcos�(✓) and tsin�+2s(✓) = tsin�(✓).

It should be noted that the formulas for tsin�(✓) and tcos�(✓) given by P and Rfrom Definition 3 are only valid for values of � in the first quadrant. Since Theorem1 gives explicit formulas for entire � and ✓ periods, we may use this theorem andthe two periodic properties stated above to give values for tsin�(✓) and tcos�(✓) forany (�, ✓) 2 R ⇥ R. Figure 4 contains a graph of tsin�✓ for two periods of � andtwo periods of ✓.

10 KELLY DELP AND MICHAEL FILIPSKI

0

1

2

3

4

f

05

1015

q

-1

0

1

Figure 4. Graph of the generalized sine function.

Table 1 contains a family of cross sections. Referring to the formulas in Theorem1 we see that for fixed � these functions are piecewise linear. We invite the interestedreader to verify that these functions are constant when ✓ = 2n for some interger n.

Recall that in the Euclidean metric, sin(✓ + ⇡2 ) = cos(✓). The cross-sections for

the sine and cosine functions when � is fixed suggest a similar identity, which weprove.

Proposition 6. tsin�(✓ + 2) = tcos�(✓)

Proof. While this identity follows from the symmetry of the space, Theorem 1gives explicit formulas for tsin�✓ and tcos�✓, so we need only check the formulasto verify this identity. Assume that 0 � < 2 and �� ✓ 2� �, which implies2� � ✓ + 2 4� �. For angles in the interval [2� �, 4� �],

tsin�✓ = 1 + ↵(✓ � 2)(�� 1).

Therefore,

tsin�(✓ + 2) = 1 + ↵((✓ + 2)� 2)(�� 1)) = 1 + ↵ ✓(�� 1),

which is equal to tcos�✓, when �� ✓ 2 � �. The other intervals can beverified similarly.

⇤

4.2. Maximum and minimum values.

Theorem 2. The maximum value of tsin�✓ and tcos�✓ is

12 + 1p

2; the minimum

value is �( 12 + 1p2).

Proof. By Proposition 6, the maximum of the sine function is equal to the maximumof the cosine function. Also, by Proposition 5, the minimum of the sine functionis equal to the negative of the maximum. Therefore it is su�cient to verify themaximum of the sine function.

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 11

tsin�✓ tcos�✓ tsin�✓ tcos�✓

� = 0

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

� = .5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 1.5

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

� = .85

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 3

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

� = 1

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 4.5

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

� = 1.15

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10 15

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 6

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

� = 1.5

5 10

-1.5

-1.0

-0.5

0.5

1.0

1.5

5 10

-1.5

-1.0

-0.5

0.5

1.0

1.5

✓ = 7.5

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

Domain: �� ✓ 16� � Domain: 0 � 2

Table 1. Cross sections.

The sine function has a ✓-period of 8, and a �-period of 2. However, the maximumof the sine function must occur when the sine is positive, hence ✓ must be in theinterval [0, 4], by Proposition 3. It is therefore su�cient to find the maximum oftsin�✓ on the region defined by 0 � 2 and 0 ✓ 4. We will use standardtechniques from multivariable calculus to maximize this function.

As tsin�✓ is piecewise defined, we will consider the intervals: [0, 2��], [2��, 4��]and [4� �, 4]. Recall that

↵ =1

�2 � 2�+ 2=

1

(�2 � 1) + 1,

which is positive for all �. When ✓ is in the interval [0, 2� �], tsin�✓ = ↵✓, and ✓in [4� �, 4] implies tsin�✓ = ↵(4� ✓). The partial derivatives with respect to ✓ ofthese functions are ↵ and �↵, therefore tsin�✓ is increasing with respect to ✓ on[0, 2� �], and decreasing in ✓ on [4� �, 4]. This implies the absolute maximum oftsin�✓ occurs when ✓ is in the middle interval.

When 2� � ✓ 4� �,

tsin�✓ = 1 +(✓ � 2) (�� 1)

�2 � 2�+ 2.

12 KELLY DELP AND MICHAEL FILIPSKI

The partial derivatives are,

@

@�

1 +

(✓ � 2) (�� 1)

�2 � 2�+ 2

�=

(2�� �2)(✓ � 2)

(�2 � 2�+ 2)2

and

@

@✓

1 +

(✓ � 2) (�� 1)

�2 � 2�+ 2

�=

�� 1

�2 � 2�+ 2.

These are both zero only when (�, ✓) = (1, 2). In this case, tsin1(2) = 1. Wenow check the boundary conditions.

When � = 0, 2 ✓ 4, and tsin�✓ = 2 +�✓

2, which has a maximium of 1. Note

that tsin�✓ has the same maximum when � = 2 because of the �-periodic propertypreviously stated.

When ✓ = 2� �, we have

g(�) = tsin�(2� �) = 1� (�� 1)�

�2 � 2�+ 2.

The derivative of this function is

g0(�) =�2 � 4�+ 2

(�2 � 2�+ 2)2.

This function is zero when � = 2±p2. Only one of these values, � = 2�

p2, is

in the region under consideration. For this value of �, we have ✓ =p2 and we see

the value of the sine function is

tsin2�p2

p2 =

1

2+

1p2.

When ✓ = 4� �, we have

h(�) = tsin�(4� �) = 1� (�� 2)(�� 1)

�2 � 2�+ 2.

The derivative of this function is

h0(�) =2� �2

(�2 � 2�+ 2)2

For values of � in the interval [0, 2], this derivative is zero when � =p2. Then

✓ = 4�p2, and

tsinp2(4�p2) =

1

2+

1p2.

We can therefore conclude for values in the region 0 � 2, and 0 ✓ 4, the function tsin�✓ achieves its absolute maximum, 1

2 + 1p2, in two locations:

(2�p2,p2) and (

p2, 4�

p2).

⇤

Corollary 1. The hypotenuse of a right triangle in taxicab space is not always thelongest side of the triangle.

TWO-PARAMETER TAXICAB TRIG FUNCTIONS 13

References

[1] Ziya Akca and Rustem Kaya. On the taxicab trigonometry. J. Inst. Math. Comput. Sci. Math.

Ser., 10(3):151-159, 1997.[2] Javier Alonso, Horst Martini, and Senlin Wu. On Birkho↵ orthogonality and isosceles orthog-

onality in normed linear spaces. Aequat. Math., 83(1-2):153-189, 2012.[3] Juan Carlos Alvarez Paiva and Anthony Thompson. On the Perimeter and Area of the Unit

Disc. Amer. Math. Monthly, 112(2):141-154, 2005.[4] J.C. Alvarez Paiva and A.C. Thompson. Volumes on normed and Finsler spaces. A sampler of

Riemann-Finsler geometry, Math. Sci. Res. Inst. Publ., 50:1-48, Cambridge Univ. Press, 2004.[5] Ruth Brisbin and Paul Artola. Taxicab trigonometry. Pi Mu Epsilon Journal, 8(2):89-95, 1985.[6] Doris J. Schattschneider. The Taxicab Group. Amer. Math. Monthly, 91(7):423-428, 1984.[7] A. C. Thompson. Minkowski Geometry. Encyclopedia of Mathematics and its Applications,

63. Cambridge University Press, 1996.[8] Kevin Thompson, Tevian Dray. Taxicab Angles and Trigonometry. Pi Mu Epsilon Journal,

11(2):87-96, 2000.[9] Nico Duvelmeyer. Angle measures and bisectors in Minkowski planes. Canad. Math. Bull.,

48(4):523-534, 2005.

Current address: Department of Mathematics, Ithaca College, Ithaca, NY 14850.E-mail address: kelly.delp@gmail.com

Current address: Department of Mathematics, Bu↵alo State College, Bu↵alo, NY 14222.E-mail address: filipsmg01@mail.buffalostate.edu